Solution:

From the figure

(i) ABCD is a cyclic quadrilateral

Ext. ∠DCE = ∠BAD

∠BAD = x^o

Now arc BD subtends ∠BOD at the center

And ∠BAD at the remaining part of the circle.

∠BOD = 2 ∠BAD = 2 x

2 x = 150^o (x = 75⁰)

(ii) ∠BCD + ∠DCE = 180^o

(Linear pair)

∠BCD + 80⁰ = 180^o

∠BCD = 180⁰ – 80⁰ = 100^o

Arc BAD subtends reflex ∠BOD at the

Centre and ∠BCD at the remaining part of the circle

Reflex ∠BOD = 2 ∠BCD

X^o = 2 × 100^o = 200^o

(iii) In ∆ACB,

∠CAB + ∠ABC + ∠ACB = 180^o

(Angles of a triangle)

But
∠ACB = 90^o

((Angles of a semicircle)

25^o + 90^o + ∠ABC = 180^o

=115^o + ∠ABC = 180^o

∠ABC = 180^o – 115⁰=65^o

ABCD is a cyclic quadrilateral

∠ABC + ∠ADC = 180^o

(Opposite angles of a cyclic quadrilateral)

65^o + x^o =180^o

x^o = 180^o -65^o = 115^o